# QUESTIONS ON SIMPLE INTEREST (PART-I)

## QUESTIONS ON SIMPLE INTEREST (PART-I)

#### QUERY 1

**Ramesh borrows Rs 7000 from a bank at SI. After 3 years he paid Rs 3000 to the bank and at the end of 5 yrs from the date of borrowing he paid Rs 5450 to the bank to settle the account. Find the rate of interest?**

A) 2%

B) 3%

C) 4%

D) 5%

**Suryadipt singh**

let the rate be r%.

Therefore interest on 7000 for first 3 yrs = 210r

Principal after 3 yrs = 7000 – 3000 = 4000

Interest on 4000 for last 2 years @ r% = 80r

Therefore the total interest =210r + 80r = 290r

Now the total of interest actually paid = (3000 + 5450) – 7000 = 1450

Hence 290r = 1450

r = 5% (option ‘D’)

#### QUERY 2

**Two equal sums of money were lent out at 6% and 8% simple interest for 4 yrs. The total interest earned was Rs. 11,200. What was the total sum?**

A) 40,000

B) 80,000

C) 20,000

D) 50,000

**RONNIE BANSAL**

Let those equal sums of money be Rs 100 each

Then the interest @6% for 4 years = 24

and @8% = 32

Thus total of interest 24 + 32 = 56

But this interest is on 100 + 100 = 200

So by unitary method the total sum = (200 x 11200)/56 = 40000 (option ‘A’)

**JAYANTCHARAN CHARAN**

Since the sums lent were equal we can safely take the average interest per year.

Now the average of interest = (6+8)/2 = 7. So the interest for 4 = 7*4 = 28

Means 28% of something is 11200

Therefore the total sum = (11200*100)/28 = 40000 (option ‘A’)

#### QUERY 3

**A certain sum of money at simple interest amounts to 1012 in 2.5 years and 1067.2 in 4 years. The rate of interest per annum is?**

A) 2.5

B) 3

C) 4

D) 5

**RONNIE BANSAL**

Simple interest for 1.5 years on the money lent = 1067.20 – 1012.00 = 55.20

Therefore interest for one year = (2/3)*55.20 = 36.80

And for 2.5 years = 2.5*36.80 = 92

Now it becomes simple to think that Rs 1012 is including the interest of Rs 92; so the principal = 1012 – 92 = 920

Now you can say the interest on 920 for 1 year is 36.80

Hence the rate of interest = (36.80*100)/920 = 4 (option ‘C’)

#### QUERY 4

**A 10 years old boy got Rs 50,000 from his father, which he deposited in a trust. The trust invested this money at 4% annual interest. This trust gives Rs 1,200 per year to the boy and the expenses of trust is Rs 300 per year. How much money will this boy get when he turns 18 years old?**

A) 54000

B) 50000

C) 58800

D) 54200

**RONNIE BANSAL
**Interest per year = 4% of 50,000 = 2,000

Total sum payable to the boy after 8 (18 – 10) years = 50,000 + Interest for 8 years = 50,000 + 16,000 = 66,000

Already paid to him by the trust = 8 (paid in cash per year + annual expenses) = 8 (1,200 + 300) = 8 x 1,500 = 12,000

Therefore money to be got by that boy on completion of 18 years of age = 66,000 – 12,000 = 54,000 (option ‘A’)

#### QUERY 5

**Rs 3000 amounts to Rs 6750 after a certain period of time when the interest is applied annually. What will the amount be if the time period is taken half of the previous time?**

A) Rs 3750

B) Rs 4000

C) Rs 4550

D) Rs 4875

**MAHA GUPTA
**There is no mention of compound interest having been the basis. When such is the case it’s simple interest taken into consideration. If so the solution should be:

Interest for full time = 6750 – 3000 = 3750

Interest for half the time = 3750/2 = 1875

Therefore the amount = 3000 + 1875 = 4875 (option ‘D’)

#### QUERY 6

**If a sum of Rs 12,000 is invested in two parts, the first part for 5 years @8% p.a. and the second one for 4 years @10% p.a. If SI of first part is three times of second part, find the difference in those parts.**

A) 6000

B) 5000

C) 5500

D) 6500

**MAHA GUPTA**

Let x be the first part; obviously the second part then is 12,000 – x

As the interest of the first part is thrice of the second x*5*8/100 = 3*[(12,000 – x)*4*10

=> x = 9,000 i.e. the first part

Therefore the second part = 12,000 – 9,000 = 3,000

But we have to find the difference

Therefore the difference = 9,000 – 3,000 = 6,000 (option ‘A’)

ANOTHER METHOD

Let the amount invested in each part = Rs 100

Therefore the simple interest on the first part = 100*5*8/100 = 40

And interest on the second part = 100*4*10 = 40

Therefore ratio of interest earned on equal amount = 40 : 40 = 1 : 1

And the ratio of amount to be invested to get equal interest = inverse ratio of the above i.e. 1: 1

Hence to get 3 times interest on the first part, the amount must be divided in the ratio

Therefore ratio of the two amounts invested = (1*3) ; (1*1) = 3 : 1

Hence the two parts 12,000*3/4 = 9,000 (first part); and 12,000*1/4 = 3,000 (second part)

So the difference = 9,000 – 3,000 = 6,000 (option ‘A’)

TRICK

**JAYANT CHARAN**

You can solve it orally also. Since 5*8 i.e. 40 is equal to 4*10 which is also 40, and SI of the first part is three times of the second; means one part is thrice the other part in this case. So 12,000 will be broken into 4 parts, one of them being 9,000 and the other being 3,000.

Hence the difference = 9,000 – 3,000 = 6,000 (option ‘A’)

#### QUERY 7

**A certain sum of money amounted Rs 825 at 7% in a time in which Rs 560 amounted to Rs 960 at 6%. If the rate of interest is simple, find the sum.**

A) 450

B) 475

C) 500

D) 600

**MAHA GUPTA**

Interest on 560 for 1 year at 6% = 33.60

Total interest on 560 = 960 – 560 = 400

Therefore number of years for which it was lent/borrowed = 400/33.60 = 250/21 years

Now let the principal = x

So, x + x*(7/100)*(250/21) = 825

=> 550x/300 = 825

=> x = 825*(300/550)

=> x = 450

Therefore the sum lent/borrowed = 450 (option ‘A’)

**NOTE:** With answer options it can be done in quicker time.

#### QUERY 8

**A man deposited Rs 400 for 2 years, Rs 550 for 4 years and Rs 1200 for 6 years. He received Rs 1020 as the total simple interest. The rate of interest per annum is?**

A) 7%

B) 7.5%

C) 7.25%

D) 10%

**MAHA GUPTA**

In such a case better to find 1 year equivalent sum; and then do calculations taking the period as 1 year. See how

1 year equivalent sums

Case-I = 400*2 = 800

Case-II = 550*4 = 2200

Case-III = 1200*6 = 7200

Total = (800 + 2200 + 7200) = 10200

We know INTEREST = PRT/100

You’ll have to remember that TIME here is 1 year as our sum now is 1 year equivalent

So, 1020 = (10200*R*1)/100

=> R = 102000/10200 = 10% (option ‘D’)

#### QUERY 9

**Simple interest on a certain sum of money at 4% per annum for 4 years is Rs 80 more than the interest on the same sum for 3 years at 5% per annum. Find the sum.**

A) 10000

B) 12000

C) 8000

D) 10500

**MAHA GUPTA**

Let the sum be 100

Therefore the difference of interest in the two given cases = [^{(100*4*4)}⁄_{100} – ^{(100*5*3)}⁄_{100}] = 16 – 15 = 1

Now if the difference is 1 the sum = 100

If the difference is 80 the sum = 100*80 = 8000 (option ‘C’)

#### QUERY 10

**The interest of one rupee in one month is 1 paise. What is the rate of interest?**

A) 10%

B) 12%

C) 10.75%

D) 11%

**MAHA GUPTA**

Interest on 1 rupee in 1 year (12 months) = 1*12 = 12 paise

So interest on 100 rupees in 1 year = 12*100 = 1200 paise = 12 rupees

Hence the rate of interest = 12% (option ‘B’)